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Research article

Eventual smoothness of generalized solutions to a singular chemotaxis system for urban crime in space dimension 2


  • Received: 08 February 2023 Revised: 15 March 2023 Accepted: 20 March 2023 Published: 28 March 2023
  • This paper is concerned with a chemotaxis system in a two-dimensional setting as follows:

    {ut=Δuχ(ulnv)κuv+ruμu2+h1,vt=Δvv+uv+h2,

    with the parameters χ,κ,μ>0 and rR, and with the given functions h1,h20. This model was originally introduced by Short et al for urban crime with the particular values χ=2,r=0 and μ=0, and the logistic source term ruμu2 was incorporated into () by Heihoff to describe the fierce competition among criminals. Heihoff also proved that the initial-boundary value problem of () possesses a global generalized solution in the two-dimensional setting. The main purpose of this paper is to show that such a generalized solution becomes bounded and smooth at least eventually. In addition, the long-time asymptotic behavior of such a solution is discussed.

    Citation: Zixuan Qiu, Bin Li. Eventual smoothness of generalized solutions to a singular chemotaxis system for urban crime in space dimension 2[J]. Electronic Research Archive, 2023, 31(6): 3218-3244. doi: 10.3934/era.2023163

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  • This paper is concerned with a chemotaxis system in a two-dimensional setting as follows:

    {ut=Δuχ(ulnv)κuv+ruμu2+h1,vt=Δvv+uv+h2,

    with the parameters χ,κ,μ>0 and rR, and with the given functions h1,h20. This model was originally introduced by Short et al for urban crime with the particular values χ=2,r=0 and μ=0, and the logistic source term ruμu2 was incorporated into () by Heihoff to describe the fierce competition among criminals. Heihoff also proved that the initial-boundary value problem of () possesses a global generalized solution in the two-dimensional setting. The main purpose of this paper is to show that such a generalized solution becomes bounded and smooth at least eventually. In addition, the long-time asymptotic behavior of such a solution is discussed.



    The importance of existence and uniqueness theorems for initial value and boundary value problems (IVP and BVP) involving the classical derivative operator is indisputable because, without them, one cannot understand modeled systems correctly and make predictions how they behave. Recently, with the popularity of fractional derivative operators such as Riemann-Liouville (R-L), Caputo (C), etc., the equations involving these operators have begun to be studied in detail (See, [1,2,3,4,5,6]). However, such a generalization leads to some difficulties and differences especially in R-L case. For instance, unlike the initial value problems involving the classical derivative, the existence of continuous solution to some IVPs in the sense of R-L derivative strictly depends on the initial values and smoothness conditions on nonlinear functions in right-hand side of equations in IVPs. For example, as shown in [7], the following initial value problem for R-L derivative of order σ(0,1)

    {Dσω(x)=f(x,ω(x)),x>0ω(0)=b0 (1.1)

    has no continuous solution when f is continuous on [0,T]×R. Therefore, this equation was investigated when subjected to different initial values (See, Theorem 3.3 in [1]). By considering the aforementioned remark, we investigate the existence and uniqueness of solutions to the following problem

    {Dσω(x)=f(x,ω(x),Dσ1ω(x)),x>0ω(0)=0,Dσ1ω(x)|x=0=b, (1.2)

    where σ(1,2), bR, Dσ represents the Riemann-Liouville fractional derivative of order σ, which is given by

    Dσω(x)=1Γ(2σ)d2dx2x0ω(t)(xt)σ1dt,

    and f fulfills the following condition:

    (C1) Let f(x,t1,t2)C((0,T]×R×R) and xσ1f(x,t1,t2)C([0,T]×R×R),

    where C(X) represents the class of continuous functions defined on the subspace X of R3.

    The equation in (1.2) was first considered by Yoruk et. al. [8], when the second initial value is also homogeneous (b=0) and the right-hand side function is continuous on [0,T]×R×R. They gave Krasnoselskii-Krein, Roger and Kooi-type uniqueness results. As seen from condition (C1), the equation in (1.2) has a singularity at x=0. Such singular equations involving R-L and Caputo derivatives were recently investigated in [7,9,10,11,12] and they proved local or global existence theorem for initial value problems involving singular equations. In the investigation of existence of solutions to these problems, converting the problem into the Volterra-type integral equation is one of the referenced tools. These integral equations have weakly and double singularities since the corresponding fractional differential equations are singular. For this reason, some new techniques or lemmas had to be developed to reveal the existence and uniqueness of solutions to integral equations (See for example [10,11]). In this paper, we also encounter a Volterra-type integral equations having a single singularity when we transform the problem (1.2) into a integral equation. For the uniqueness of solutions to the integral equation we need to generalize the definitions and methods previously given in [7,8,13] and use the tools of Lebesgue spaces such as Hölder inequality. These uniqueness theorems are the type of Nagumo-type, Krasnoselskii-Krein and Osgood which are well-known in the literature (See, [13,14,15,16]) Moreover, we give a Peano-type existence theorem for this problem as well.

    We first give definition of the space of functions where we investigate the existence of solutions to the problem (1.2). The space is defined below as in [17] :

    Theorem 2.1. The space of continuous functions defined on [0,T], whose R-L fractional derivative of order σ1, 1<σ<2 are continuous on [0,T] is a Banach space when endowed with the following norm:

    ||ω||σ1=||ω||+||Dσ1ω||,

    where ||.|| is the supremum norm defined on the class of continuous functions. This space will be denoted by Cσ1([0,T]).

    According to this space, we define R-L integral and R-L derivative of higher order [11]. The lower terminal points of integrals in their formulas will be taken as zero.

    Definition 2.1. The R-L integral of order σ>0 of a function ω(x)C0[0,T] is defined for all x[0,T] by

    Iσω(x):=1Γ(σ) x0ω(t)(xt)1σdt. (2.1)

    Definition 2.2. For 1<σ<2 and ω(x)Cσ1[0,T] with Dσ1ω(x)C1(0,T]L1[0,T] the R-L fractional derivative Dβω is defined for all x(0,T] by

    Dσω(x)=1Γ(2σ)d2dx2x0ω(t)(xt)σ1dt. (2.2)

    The local existence of solutions to problem (1.2) will be proved with the aid of Schauder fixed point theorem [18]:

    Theorem 2.2. Let C be a closed, bounded, convex subset of a Banach space X:={u:IR continuous:IR closed  and  bounded  interval}. If operator S:CC is continuous and, if S(C) is an equicontinuous set on I, then S has at least one fixed point in C.

    The one of mathematical tools used for showing the existence and uniqueness of the desired type of solution to a given initial or boundary value problem is first to convert them into an integral equation. One investigates the existence and uniqueness of the solution to the integral equation instead of the associated problem. Here, we follow this way by taking the aid of the lemma given below:

    Lemma 3.1. Under condition (C1), if ωCσ1[0,T] is a solution of problem (1.2), then ωCσ1[0,T] is a solution of the following integral equation

    ω(x)=bΓ(σ)xσ1+1Γ(σ)x0f(t,ω(t),Dσ1ω(t))(xt)1σdt (3.1)

    and, vice versa.

    Proof. We assume that ωCσ1[0,T] is a solution of problem (1.2). By condition (C1), we have f(x,ω(x),Dσ1ω(x)) is continuous on (0,T] and xσ1f(x,ω(x),Dσ1ω(x)) is continuous on [0,T]. It means that f(x,ω(x),Dσ1ω(x)) is integrable, i.e f(x,ω(x),Dσ1ω(x))C(0,T]L1[0,T]. Then, by integrating the both sides of the equation in (1.2) and using the relation IDσ=IDDσ1

    Dσ1ω(x)=Dσ1ω(0)+If(x,ω(x),Dσ1ω(x))C[0,T]

    is obtained. From here, by integration of the both sides of last equation and by use of IDσ1=IDI2σ we have

    I2σω(x)=bx+I2f(x,ω(x),Dσ1ω(x))C1[0,T] (3.2)

    where we used Dσ1ω(0)=b and I2σω(0)=0 since ω(0)=0.

    If the operator Iσ1 is applied to the both sides of (3.2), then by semigroup and commutative property of R-L derivative we get

    Iω(x)=bxσΓ(1+σ)+I[Iσf(x,ω(x),Dσ1ω(x))] (3.3)

    for all x[0,T]. Differentiating the both sides of (3.3), we have

    ω(x)=bxσ1Γ(σ)+Iσf(x,ω(x),Dσ1ω(x))

    which is the equivalent to the integral Eq (3.1).

    Now we suppose that ωCσ1[0,T] is a solution of integral Eq (3.1), and let us show that ω is a solution of the problem (1.2). If Dσ is applied to the both sides of (3.1), and then, if

    DσIσω(x)=ω(x)  for all  ωCσ1[0,T]

    is used, then one can observe that ωCσ1[0,T] satisfies the equation in (1.2). Moreover, let us prove that ωCσ1[0,T] also fulfills initial value conditions. By change of variables and condition (C1) we have

    ω(0)=limx0+ω(x)=bΓ(σ)xσ1+1Γ(σ)limx0+x0f(t,ω(t),Dσ1ω(t))(xt)1σdt=1Γ(σ)limx0+x0tσ1f(t,ω(t),Dσ1ω(t))tσ1(xt)1σdt=1Γ(σ)limx0+x10(xτ)σ1f(xτ,ω(xτ),Dσ1ω(xτ))τσ1(1τ)1σdτ=0, (3.4)

    since the integral is finite. Thus ω satisfies the first initial condition in (1.2).

    Now let us show that ω provides the second initial condition in (1.2). If Dσ1 is applied to both sides of (3.1), and if the relation Dσ1Iσh(x)=Ih(x) is used, then we can first get

    Dσ1ω(x)=b+x0f(t,ω(t),Dσ1ω(t))dt. (3.5)

    From here, by passing the limit as x0+, we then obtain

    Dσ1ω(0)=b+limx0+x0f(t,ω(t),Dσ1ω(t))dt=b+limx0+x01tσ1tσ1f(t,ω(t),Dσ1ω(t))dt=b+limx0+x2σ101τσ1(xτ)σ1f(xτ,ω(xτ),Dσ1ω(xτ))dτ=b, (3.6)

    since 2σ>0 and the integral is finite due to the continuity of tσ1f(t,ω(t),Dσ1ω(t)) on [0,T]. Consequently, it has been shown that any solution of (3.1) provides the problem (1.2) if condition (C1) is assumed to be satisfied.

    Theorem 3.1 (Existence). Let condition (C1) be satisfied, and assume that there exist positive real numbers r1, r2 and M such that |xσ1f(x,ω,v)|Mfor  all(x,ω,v)I=[0,T]×[r1,r1]×[br2,b+r2]. Then problem (1.2) admits at least one solution in Cσ1[0,T0 ], where

    T0 ={TifT<rC(b,σ,M)rC(b,σ,M)ifTrC(b,σ,M)1[rC(b,σ,M)]σ1,ifTrC(b,σ,M),1rC(b,σ,M)and1<σ1.5[rC(b,σ,M)]2σ,ifTrC(b,σ,M),1rC(b,σ,M)and1.5<σ<2, (3.7)

    and

    r=r1+r2andC(b,σ,M)=[|b|Γ(σ)+M(1+Γ(3σ)2σ)]. (3.8)

    Proof. As it is known from Lemma 3.1, solutions of problem (1.2) are solutions of integral equation (3.1) as well. Moreover, the fixed points of the opeator S:Cσ1[0,T0 ]Cσ1[0,T0 ] defined by

    Sω(x)=bΓ(σ)xσ1+x0f(t,ω(t),Dσ1ω(t))(xt)1σdt (3.9)

    interfere with solutions of the integral equation. For this reason, it is sufficient to prove that operator S admits at least one fixed point. For this, it will be verified that operator S satisfies the hypotheses of Schauder fixed-point theorem. Let us start with showing the following inclusion to be valid:

    S(Br)Br

    where

    Br={ωCσ1[0,T0 ]:||ω||+||Dσ1ωb||r}

    is a closed compact subset of Cσ1[0,T0 ]. Accordingly to the norm on Cσ1[0,T0 ], upper bounds of Sω(x) and Dσ1Sω(x)b can be determined as follows:

    |Sω(x)||b|Γ(σ)xσ1+1Γ(σ)x0|tσ1f(t,ω(t),Dσ1ω(t)|tσ1(xt)1σdt|b|Γ(σ)xσ1+MΓ(σ)10xτσ1(1τ)1σdτ|b|Γ(σ)xσ1+Γ(2σ)Mx (3.10)

    and

    |Dσ1Sω(x)b|x0|tσ1f(t,ω(t),Dσ1ω(t)|tσ1dtMx2σ10τ1σdτ=Mx2σ2σ. (3.11)

    From (3.10)) and (3.11),

    |Sω(x)|+|Dσ1Sω(x)b||b|Γ(σ)xσ1+Γ(2σ)Mx+Mx2σ2σ (3.12)

    is obtained. Taking supremum over [0,T0] for a T0>0 for the right hand-side of the above equation,

    |Sω(x)|+|Dσ1Sω(x)b|C(b,σ,M)Tα0 (3.13)

    can be written, where αΩ={σ1,1,2σ}. α depends on values of b,M,σ,r. To determine T0 and α, let

    C(b,σ,M)Tα0=r.

    If Tα0=rC(b,σ,M)<1, then it is observed that T0<1 for any αΩ. If Tα0=rC(b,σ,M)1, it must be T01 for any αΩ. Thus,

    supx[0,T0][|Sω(x)|+|Dσ1Sω(x)b|]C(b,σ,M)Tα0=r, (3.14)

    where

    T0:=[rC(b,σ,M)]1/α

    and

    α={1ifrC(b,σ,M)1σ1ifrC(b,σ,M)<1and1<σ1.52σifrC(b,σ,M)<1and1.5σ<2. (3.15)

    As a result, for all cases we obtain

    ||Sω||+||Dσ1Sωb||r,

    which is the desired.

    Now, let us prove the equicontinuity of S(Br)Cσ1[0,T0 ]. Since the composition of uniformly continuous functions is so as well, the function xσ1f(x,ω(x),Dσ1ω(x)) is uniformly continuous on [0,T0 ]. Because for any ωBr, both ω(x) and Dσ1ω(x) and xσ1f(x,ω,v) are uniformly continuous on I, respectively. Therefore, for given any ϵ>0, one can find a δ=δ(ϵ)>0 so that for all x1,x2[0,T0 ] with |x1x2|<δ it is

    |xσ11f(x1,ω(x1),Dσ1ω(x))xσ12f(x2,ω(x2),Dσ1ω(x2))|<Kϵ,

    where K=max(1T0 Γ(2σ),2σT0 2σ). It follows that

    |Sω(x1)Sω(x2)|+|Dσ1Sω(x1)Dσ1Sω(x2)|10|h(ηx1)h(ηx2)|Γ(σ)η1σ(1η)σ1xdη+10|h(ηx1)h(ηx2)|η1σx2σdη<T0 Γ(2σ)Kϵ+T0 2σ2σKϵ=ϵ,

    where h(x)=xσ1f(x,ω(x),Dσ1ω(x)). This implies that S(Br) is an equicontinuous set of Cσ1[0,T0 ].

    Finally, the continuity of S on Br will be proven. Assume that {ωk}k=1Br is a sequence with ωkCσ[0,T0 ]ω as k. Then, one can easily conclude that ωk and Dσ1ωk(t) converges uniformly to ω and Dσ1ω(t), respectively. With these and the uniform continuity of xσ1f(x,ω,v) on I=[0,T]×[r1,r1]×[br2,b+r2], it leads to

    SωkSωσ1=supx[0,T0 ]|1Γ(σ)x0[f(t,ωk(t),Dσ1ωk(t))f(t,ω(t),Dσ1ω(t))](xt)1σdt|+supx[0,T0 ]|x0[f(t,ωk(t),Dσ1ωk(t))f(t,ω(t),Dσ1ω(t))]dt|supηx[0,T0 ]10(ηx)σ1|f(ηx,ωk(ηx),Dσ1ωk(ηx))f(ηx,ω(ηx),Dσ1ω(ηx))|Γ(σ)ησ1(1η)1σxdη+supηx[0,T0 ]10(ηx)σ1|f(ηx,ωk(ηx),Dσ1ωk(ηx))f(ηx,ω(ηx),Dσ1ω(ηx))|ησ1x2σdη0ask.

    In conclusion, since hypotheses of Theorem 2.2 are fulfilled, it implies that operator S admits at least one fixed point in Cσ1[0,T0 ], which is a solution of problem (1.2) as well.

    The mean value theorem for R-L derivative of order σ(0,1) was correctly given by [7]. Now, its counterparts for order σ(1,2) is given as follows:

    Lemma 3.2. Let σ(1,2) and ωCσ1([0,T]). Then, there is a function μ:[0,T][0,T] with 0<μ(x)<x so that

    ω(x)=Dσ1ω(0)xσ1Γ(σ)+Γ(2σ)x(μ(x))σ1Dσ1ω(μ(x)),

    is satisfied.

    The lemma can be proved by following the way used in [7] and so we omit it here. With the aid of this lemma we can obtain the Nagumo-type uniqueness:

    Theorem 3.2. (Nagumo type uniqueness) Let 1<σ<2, 0<T< and let condition (C1) be satisfied. Moreover, assume that there exists a positive real number L2σmax(T,T2σ)(1+Γ(3σ)) such that the inequality

    xσ1|f(x,ω1,v1)f(x,ω2,v2)|L(|ω1ω2|+|v1v2|) (3.16)

    is fulfilled for all x[0,T] and for all ωi,viR with i=1,2. Then, (1.2) has at most one solution in the space of Cσ1([0,T0]).

    Proof. We have just showed the existence of the solution to problem (1.2) in the previous theorem. For the uniqueness, we first assume that (1.2) admits two different solutions such as ω1 and ω2 in the space of Cσ1([0,T0]). Let us define a function Φ(x) to be in the form

    Φ(x):={|ω1(x)ω2(x)|+|Dσ1ω1(x)Dσ1ω2(x)|,x>0                    0           ,x=0.

    Since ω1,ω2Cσ1([0,T]), the continuity of Φ(x) on x(0,T0] can obviously be seen. For its continuity at x=0,

    0limx0+Φ(x)=limx0+1Γ(σ)|x0f(t,ω1(t),Dσ1ω1(t))f(t,ω2(t),Dσ1ω2(t))(xt)1σdt|+limx0+|x0f(t,ω1(t),Dσ1ω1(t))f(t,ω2(t),Dσ1ω2(t))dt|10limx0+x|H(xη,ω1(xη))H(xη,ω2(xη))|ησ1(1η)1σdη+10limx0+x2σ|H(xη,ω1(xη))H(xη,ω2(xη))|ησ1dη=0,

    where H(x,ω(x))=xσ1f(x,ω(x),Dσ1ω(x)) and we made the change of variable t=xη and used condition (C1), respectively. Consequently, limx0+Φ(x)=0=Φ(0).

    The fact that Φ(x)0 on [0,T] allows us to choose a point x0(0,T] so that

    0<Φ(x0)=|ω1(x0)ω2(x0)|+|Dσ1ω1(x0)Dσ1ω2(x0)|.

    By using the mean value theorem given by Lemma 3.2

    |ω1(x0)ω2(x0)|=Γ(2σ)x0|xσ11Dσ1(ω1ω2)(x1)|=Γ(2σ)x0xσ11|f(x1,ω1(x1),Dσ1ω1(x1))f(x1,ω2(x1),Dσ1ω2(x1))| (3.17)

    is obtained for x1(0,x0).

    Secondly, for the estimation of |Dσ1ω1(x0)Dσ1ω2(x0)|, we have from the well-known integral mean theorem for the classical calculus

    |Dσ1ω1(x0)Dσ1ω2(x0)|=x00tσ1|f(t,ω1(t),Dσ1ω1(t)f(t,ω2(t),Dσ1ω2(t)|tσ1dt=x2σ02σxσ12|f(x2,ω1(x2),Dσ1ω1(x2))f(x2,ω2(x2),Dσ1ω2(x2))|, (3.18)

    where x2(0,x0).

    We assign x3 as one of the points x1 and x2 so that |H(x3,ω1(x3))H(x3,ω2(x3))| :=max(|H(x1,ω1(x1))H(x1,ω2(x1))|,|H(x2,ω1(x2))H(x2,ω2(x2))|).

    Thus, from (3.17) and (3.18), we have

    0<Φ(x0)(Γ(2σ)x0+x2σ02σ)|H(x3,ω1(x3))H(x3,ω2(x3))|max(T,T2σ)(1+Γ(3σ)2σ)xσ13|f(x3,ω1(x3))f(x3,ω2(x3))|xσ13(|ω1(x3)ω2(x3)|+|Dσ1ω1(x3)Dσ1ω2(x3)|)=Φ(x3)

    since L2σmax(T,T2σ)  (1+Γ(3σ)). Repeating the same procedure for the point x3, it enables us to find some points x6(0,x3) so that 0<Φ(x0)Φ(x3)Φ(x6). Continuing in the same way, the sequence {x3n}n=1[0,x0) can be constructed so that x3n0 and

    0<Φ(x0)Φ(x3)Φ(x6)...Φ(x3n)... (3.19)

    However, the fact that Φ(x) is continuous at x=0 and x3n0 leads to Φ(x3n)Φ(0)=0, and this contradicts with (3.19). Consequently, IVP (1.2) possesses a unique solution.

    Theorem 3.3. (Krasnoselskii-Krein type uniqueness) Let 1<σ<2 and T0=min{T0 ,1}, where T0 is defined by (3.7). Let condition (C1) be fulfilled. Furthermore, suppose that there exist a L>0 and an α(0,1) so that

    x1α(σ1)|f(x,ω1,v1)f(x,ω2,v2)|Γ(σ)α(σ1)+L2(|ω1ω2|+|v1v2|) (3.20)

    holds for all x[0,T] and for all ωi,viR with i=1,2, and that there exist C>0 satisfying (1σ)(1α)L(1α)+1>0 such that

    xσ1|f(x,ω1,v1)f(x,ω2,v2)|C(|ω1ω2|α+xα(σ1)|v1v2|α) (3.21)

    holds for all x[0,T] and for all ωi,viR with i=1,2. Then, problem (1.2) has a unique solution in the space of Cσ1([0,T0]).

    Proof. As claimed in the previous theorem, we first assume that problem (1.2) has two different solutions such as ω1(x) and ω2(x) in Cσ1([0,T0]). However, by contradiction, we will show that they are indeed equal. For this, let us first define Φ1(x)=|ω1(x)ω2(x)| and Φ2(x)=|Dσ1ω1(x)Dσ1ω2(x)| and try to find estimates for each functions by using condition (C1) and inequality (3.21). Hence, we first have

    Φ1(x)1Γ(σ)x0|f(t,ω1(t),Dσ1ω1(t))f(t,ω2(t),Dσ1ω2(t))|(xt)1σdt1Γ(σ)x0tσ1|f(t,ω1(t),Dσ1ω1(t))f(t,ω2(t),Dσ1ω2(t))|tσ1(xt)1σdtCΓ(σ)x0[Φα1(x)+tα(σ1)Φα2(x)]tσ1(xt)1σdtCΓ(σ)(x0(1tσ1(xt)1σ)qdt)1/q(x0[Φα1(t)+tα(σ1)Φα2(t)]pdt)1/pCΓ(1+(1σ)q)Γ(1+(σ1)q))Γ(σ)x1/qΩ1/p(x)

    where we used Hölder inequality with q>1 satisfying (1σ)q+1>0 and p=q/(q1), and Ω(x) is defined by

    Ω(x)=x0[Φα1(t)+tα(σ1)Φα2(t)]pdt.

    From here, we have the following estimation

    Φp1(x)Cxp/qΩ(x), (3.22)

    where C is not specified here and throughout the proof. In addition to this, the upper bound for Φ2(x) can be found as follows:

    Φ2(x)x0|f(t,ω1(t),Dσ1ω1(t))f(t,ω2(t),Dσ1ω2(t))|dtx0tσ1|f(t,ω1(t),Dσ1ω1(t))f(t,ω2(t),Dσ1ω2(t))|tσ1dtCΓ(σ)x0[Φα1(x)+tα(σ1)Φα2(x)]tσ1(xt)1σdtC(x0(1tσ1)qdt)1/q(x0[Φα1(t)+tα(σ1)Φα2(t)]pdt)1/pCΓ(1+(1σ)q)Γ(2+(σ1)q))x(1+q(1σ))/qΩ1/p(x).

    From here,

    Φp2(x)Cx(1+q(1σ))p/qΩ(x) (3.23)

    is then obtained. By using estimations in (3.22) and (3.23) in the derivative of Ω(x) we have

    Ω(x)=[Φα1(x)+tα(σ1)Φα2(x)]p2p1[(Φα1(x))p+tpα(σ1)(Φα2(x))p]=2p1[(Φp1(x))α+xpα(σ1)(Φp2(x))α]2p1[Cαxαp/qΩα(x)+Cαxpα(σ1)x(1+q(1σ))αp/qΩα(x)]Cxαp/qΩα(x). (3.24)

    If we multiply the both sides of the above inequality with (1α)Ωα(x),

    (1α)Ωα(x)Ω(x)=ddx[Ω1α(x)]Cxαp/q.

    is then obtained. Integrating the both sides of the inequality over [0,x], we get

    Ω1α(x)Cx(αp+q)/q,

    since Ω(0)=0. Consequently, this leads to the following estimation on Ω(x)

    Ω(x)Cx(αp+q)/(1α)q. (3.25)

    By considering (3.22) and (3.23) together with (3.25), one can conclude that

    Φp1(x)Cxp/qΩ(x)Cxp/qx(αp+q)/(1α)pq=Cxp+q/(1α)q.

    or

    Φ1(x)Cx(p+q)/(1α)pq=x1/(1α), (3.26)

    and

    Φp2(x)Cxp(1+q(1σ))/qΩ(x)Cxp(1+q(1σ))/qx(αp+q)/(1α)pq=Cx(1α)(1σ)pq+p+q(1α)q.

    or

    Φ2(x)Cx(1σ)+p+q(1α)pq=x(1σ)+1(1α), (3.27)

    since p+qpq=1.

    Let us now define

    Ψ(x)=xLmax{Φ1(x),Φ2(x)},

    where L(1α)<1+(1σ)(1α). If Φ1(x)=max{Φ1(x),Φ2(x)}, then from (3.26) we get

    0Ψ(x)x11αL,

    or in the case of Φ2(x)=max{Φ1(x),Φ2(x)}, by the inequality (3.27) we have the following

    0Ψ(x)x(1σ)+11αL=x(1σ)(1α)L(1α)+11α.

    In both cases, Ψ(x) is continuous on [0,T0] and Ψ(0)=0. Let us now show that Ψ(x)0 on [0,T0]. For this, suppose otherwise and let Ψ(x) This means and from its continuity one can say that there exists a point so that takes its maximum value at that point. Thus, let

    By assuming

    (3.28)

    is obtained for However, we get to a contradiction.

    On the other hand, when we get

    (3.29)

    which is contraction as well.

    Consequently, vanishes identically on This gives us the uniqueness of solutions to the considered problem.

    Theorem 3.4 (Osgood-type uniqueness). Let and be defined by (3.7), and let condition (C1) be satisfied. Furthermore, suppose that the equality

    (3.30)

    is fulfilled for all and for all with where is conjugate of satisfying and

    Moreover, assume that is a continuous, non-negative and non-decreasing function in so that and it satisfies

    (3.31)

    for any Then, has a unique solution in the space

    Proof. As made in previously given uniqueness theorems, we assume that there exist two different solutions such as and to problem (1.2) in Moreover, let and At first, we get the estimation on as follows:

    (3.32)

    where we used the inequality (3.30), Hölder inequality and the assumption on respectively. From here, it follows that

    (3.33)

    Similarly to above, we have

    (3.34)

    This leads to

    (3.35)

    Now, set

    and assume that for We will show that it can not be possible under assumptions.

    From the definition of one easily conclude that for each it is and there exists a so that Then, from estimations (3.33)–(3.35) and from the fact that is non-decreasing function

    (3.36)

    is then obtained. It can be seen that Moreover, we have

    for all From this fact, for sufficiently small we have

    Furthermore, by changing variables in the above integral and by using the continuity of and , we have

    for sufficiently small with and for However, this contradicts with the assumption on given in (3.31). Consequently, for i.e.

    Remark 3.1. It must be pointed out that, as noted in Theorem 1.4.3 in [13], the condition that function is non-decreasing can be dropped.

    Example 3.1. Let us consider the following problem

    (3.37)

    with initial conditions and Let in Theorem 3.1. Then, and Hence, problem has a solution in Now, we investigate the uniqueness of the solution to the problem in Let the function in the previous theorem defined by

    (3.38)

    It is obvious that is positive for . Since for it is non-decreasing. Also,

    (3.39)

    for any which can be seen by considering the inequality in the neighborhood of and the divergence of the integral and by applying comparison test. Moreover, by using the concavity of the nonlinear function with respect to second and third variables, we have

    (3.40)

    where for and

    Hence, assumptions of Theorem 3.4 are satisfied. So, the problem has a unique solution in

    In this research, we gave some sufficient conditions for the existence and uniqueness of a problem involving a nonlinear differential equations in the sense of R-L derivative when the right-hand side function has a discontinuity at zero. We presented an example associated with two theorems. Considering the literature, these results can be generalized and improved. Besides, one can obtain another uniqueness results for this problem as well.

    There is no conflict of interest.



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